Show that an ideal $M \ne R$ in a commutative ring $R$ with unity is maximal if and only if for every $r∈R−M$, there exists $x∈R$ such that $1_R−rx∈M$

Question

Show that an ideal $M \ne R$ in a commutative ring $R$ with unity is maximal if and only if for every $r∈R−M$, there exists $x∈R$ such that $1_R−rx∈M$

Proof : Sppose that for every $r ∈ R - M$, there exists $x ∈ R$ such that $1_R - rx ∈ M$. Let $I$ be an ideal in $R$ that properly contains $M$. Then, there exists $r ∈ I - M$. By assumption, there exists $x ∈ R$ such that $1_R - rx ∈ M$. Since $M$ is an ideal, we have $rx ∈ I$, and therefore $1_R ∈ I$. This implies that $I$ is equal to $R$, so $M$ is a maximal ideal.

I could only do the converse part. Can anyone please help me?

Answer 1

I'll note that this is a very standard result, that an ideal is maximal if and only if its quotient is a field, and there is an easy proof which uses the correspondence theorem for ideals.

But I believe you want to show it straight from the definition. So let $r\in R\setminus M$. Consider the ideal generated by $M$ and $r$, which can be described as follows:

$I=\{rx+m: x\in R, m\in M\}$

It properly contains $M$, and since $M$ is maximal we have $I=R$. If so, there are some $x\in R, m\in M$ such that $rx+m=1$. The result follows.